剑指 offer 面试题精选图解 12. 矩阵中的路径

快来打我* 2023-02-15 13:55 14阅读 0赞

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作者：景禹

## 题目描述 ##

请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。例如，在下面的 3×4 的矩阵中包含一条字符串 “bfce” 的路径（路径中的字母用加粗标出）。

[["a","b","c","e"],
    ["s","f","c","s"],
    ["a","d","e","e"]]

但矩阵中不包含字符串 “abfb” 的路径，因为字符串的第一个字符 b 占据了矩阵中的第一行第二个格子之后，路径不能再次进入这个格子。

##### 示例1 #####

> 输入：board = \[\["A","B","C","E"\],\["S","F","C","S"\],\["A","D","E","E"\]\], word = "ABCCED" 输出：true

##### 示例2 #####

> 输入：board = \[\["a","b"\],\["c","d"\]\], word = "abcd" 输出：false

## 题目解析 ##

题目考察的知识点就是深度优先搜索，在一个二维矩阵中判断是否存在单词 word 的一条路径，先从二维矩阵中选择一个任意一个起点，然后从该起点进行深度优先遍历。看下图就会更清晰：

![format_png 1][]

二维矩阵在本质上你可以将其看做一个网状的图，而关于图上面的深度优先搜索，我想你应该很清楚，不论是 DFS 还是 BFS 的实现方式，最简单的就是递归，只不过对于这样一个网状图而言，我们需要将图中的每一个顶点作为起点，并根据给定的 `word` 都进行一次深度优先搜索，相比于图的一次遍历复杂了些，判断条件也相应的多了些，仔细结合视频动画和代码理解起来就会相当容易。

下面是示例一的一个具体解释：

![format_png 2][]![format_png 3][]

![format_png 4][] ![format_png 5][] ![format_png 6][] ![format_png 7][] ![format_png 8][] ![format_png 9][]

![format_png 10][]![format_png 11][]![format_png 12][]

![format_png 13][] ![format_png 14][] ![format_png 15][] ![format_png 16][]

## 动画描述 ##

## 代码实现 ##

bool hasPath(const char* matrix, int rows, int cols, const char* str)
    {
        if(matrix == nullptr || rows < 1 || cols < 1 || str == nullptr)
            return false;
    
        bool *visited = new bool[rows * cols];
        memset(visited, 0, rows * cols);
    
        int pathLength = 0;
        for(int row = 0; row < rows; ++row)
        {
            for(int col = 0; col < cols; ++col)
            {
                if(hasPathCore(matrix, rows, cols, row, col, str,
                    pathLength, visited))
                {
                    return true;
                }
            }
        }
    
        delete[] visited;
    
        return false;
    }
    
    bool hasPathCore(const char* matrix, int rows, int cols, int row,
        int col, const char* str, int& pathLength, bool* visited)
    {
        if(str[pathLength] == '\0')
            return true;
    
        bool hasPath = false;
        if(row >= 0 && row < rows && col >= 0 && col < cols
            && matrix[row * cols + col] == str[pathLength]
            && !visited[row * cols + col])
        {
            ++pathLength;
            visited[row * cols + col] = true;
    
            hasPath = hasPathCore(matrix, rows, cols, row, col - 1,
                str, pathLength, visited)
                || hasPathCore(matrix, rows, cols, row - 1, col,
                    str, pathLength, visited)
                || hasPathCore(matrix, rows, cols, row, col + 1,
                    str, pathLength, visited)
                || hasPathCore(matrix, rows, cols, row + 1, col,
                    str, pathLength, visited);
    
            if(!hasPath)
            {
                --pathLength;
                visited[row * cols + col] = false;
            }
        }
    
        return hasPath;
    }

## 时间复杂度 ##

不论如何都要遍历二维矩阵一遍，即，对二维矩阵中的每一个顶点作为起点，最深的路径为 `word` 中单词的长度，所以最坏情况下的时间复杂度为.

## 知识点 ##

深度优先搜索 DFS ，递归

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